Điều kiện $\sin 2x \ne 0 \Leftrightarrow x\ne \dfrac{k\pi}{2};\ k\in Z$PT$ \Leftrightarrow \dfrac{\sin^4 x+\cos^4 x}{\sin 2x}=\dfrac{1}{2}. (\dfrac{\sin^2 x +\cos^2 x}{\sin x \cos x})$$\Leftrightarrow \sin^4 x +\cos^4 x = 1$ $\Leftrightarrow 1-2sin^{2}x.cos^{2}x=1$$\Leftrightarrow sin^{2}2x=0$$\Leftrightarrow sin2x=0$
Điều kiện $\sin 2x \ne 0 \Leftrightarrow x\ne \dfrac{k\pi}{2};\ k\in Z$PT$ \Leftrightarrow \dfrac{\sin^4 x+\cos^4 x}{\sin 2x}=\dfrac{1}{2}. (\dfrac{\sin^2 x +\cos^2 x}{\sin x \cos x})$$\Leftrightarrow \sin^4 x +\cos^4 x = 1$ $\Leftrightarrow 1-2sin^{2}x.cos^{2}x=1$$\Leftrightarrow sin^{2}2x=0$
Điều kiện $\sin 2x \ne 0 \Leftrightarrow x\ne \dfrac{k\pi}{2};\ k\in Z$PT$ \Leftrightarrow \dfrac{\sin^4 x+\cos^4 x}{\sin 2x}=\dfrac{1}{2}. (\dfrac{\sin^2 x +\cos^2 x}{\sin x \cos x})$$\Leftrightarrow \sin^4 x +\cos^4 x = 1$ $\Leftrightarrow 1-2sin^{2}x.cos^{2}x=1$$\Leftrightarrow sin^{2}2x=0$
$\Leftrightarrow sin2x=0$